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For what values of $a$ does the following integral converge: $$\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx$$

We have to study the integral at $0$ and $\infty$: $$\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx = \int_0^1 \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx + \int_1^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx$$ I've started checking for $a>0$: For the second integrand we have equivalence
$$\frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}} \sim \frac{x^{-2a}}{\sqrt{x^a+x^{-a}}} \sim \frac{1}{x^{5a/2}}, \ \ x \to \infty$$ So, $$\int_1^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx \sim \int_1^\infty \frac{1}{x^{5a/2}}dx $$ which converges when $5a/2>1$ or $a>2/5$.

But this approach seems to be lots of work to study fully, not to mention that I don't feel I can deal with all of the cases...

Any help is appreciated.

VIVID
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  • What is "lots of work" ? Ten minutes ? –  Jun 17 '20 at 12:43
  • You're on the right track, you just need to keep at it for the other three parts of the problem. No particular surprises occur except in regards to how you handle the log when its argument is close to 1 vs when its argument is large. – Ian Jun 17 '20 at 13:00
  • @lan yeah, what should I do with $\ln$ when the argument is large? What is it equivalent to? – VIVID Jun 17 '20 at 13:11
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    For $x \gg 1$, $\ln(1+x) \sim \ln(x)$ and that's all you can do; you have to do the analysis with this actually there. But it will help to keep in mind that $\ln(x) \ll x^\epsilon$ for any $\epsilon>0$, so if you know that $\int_1^\infty x^{a+\epsilon} dx$ converges then you know that $\int_1^\infty x^a \ln(x) dx$ also converges. – Ian Jun 17 '20 at 13:34

1 Answers1

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  1. $a > 0$

Since $$\lim_{x\to 0^{+}} \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} = 0,$$ $\int_0^1 \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges.

By using $\frac{t}{t+1} \le \ln (1+t) \le t, \ \forall t \ge 0$, we have $$ \frac{1}{4x^{5a/2}}\le \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \le \frac{1}{x^{5a/2}}, \ x \ge 1.$$ Thus, $\int_1^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges if and only if $a > \frac{2}{5}$.

As a result, $\int_0^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges if and only if $a > \frac{2}{5}$.

  1. $a < 0$

Since $$\lim_{x\to 0^{+}} \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} = 0,$$ $\int_0^1 \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges.

It is easy to obtain $$-a x^{a/2}\ln x \le \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \le -2a x^{a/2}\ln x + x^{a/2}\ln 2, \ x \ge 1.$$ Thus, $\int_1^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges if and only if $a < -2$.

As a result, $\int_0^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges if and only if $a < -2$.

River Li
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